Problem: Multiply the following complex numbers: $({-2+3i}) \cdot ({2+2i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-2+3i}) \cdot ({2+2i}) = $ $ ({-2} \cdot {2}) + ({-2} \cdot {2}i) + ({3}i \cdot {2}) + ({3}i \cdot {2}i) $ Then simplify the terms: $ (-4) + (-4i) + (6i) + (6 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -4 + (-4 + 6)i + 6i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -4 + (-4 + 6)i - 6 $ The result is simplified: $ (-4 - 6) + (2i) = -10+2i $